question_answer

In a trapezium ABCD, if AB||CD, then \[A{{C}^{2}}+B{{D}^{2}}\] equal to

A)  \[B{{C}^{2}}+A{{D}^{2}}+2AB\cdot CD\]

B)  \[A{{B}^{2}}+C{{D}^{2}}2AD\cdot BC\]

C)  \[A{{B}^{2}}+C{{D}^{2}}+2AB\cdot CD\]

D)  \[B{{C}^{2}}+A{{D}^{2}}+2BC\cdot AD\]

Correct Answer: A

Solution :

In \[\Delta ABD,\]\[\angle A\]is acute

So,       \[B{{D}^{2}}=A{{D}^{2}}+A{{B}^{2}}-2AB\cdot AQ\]       ...(i)

In\[\Delta ABC,\]\[\angle B\]is cute

So,       \[A{{C}^{2}}=B{{C}^{2}}+A{{B}^{2}}-2AB\cdot AD\]       ...(ii)

Adding Eqs. (i) and (ii), we get

\[\therefore \]\[A{{C}^{2}}+B{{D}^{2}}=(B{{C}^{2}}+A{{D}^{2}})+2AB(AB-BP-AQ)\]

\[=(B{{C}^{2}}+A{{D}^{2}})+2AB\cdot PQ\]

\[=B{{C}^{2}}+A{{D}^{2}}+2AB\cdot CD\] \[(\because PQ=DC)\]